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azamat
3 years ago
13

Which of the following is an example of a force? A. electrons moving toward a positive charge B. a child kicking a soccer ball C

. air pushing against a falling object D. all of the above
Physics
2 answers:
Andre45 [30]3 years ago
4 0

Answer:

option (d)

Explanation:

A. When an electron is moving towards a positive charge, it means there is a force of attraction between electron and positive charge.

B. When a child kicks a football, it means he is applying a force on the ball.

C. When air pushes against the falling object, it means there is an opposing force acting on the object.

Lisa [10]3 years ago
3 0
I believe it would be d. all of the above!
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An airplane is flying at 635 km per hour at an altitude of 35,000 m. What is its velocity?
Elden [556K]

Distance 350 Km

Time 1 hour

Velocity = 350 : 1 =

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your answer is a

5 0
3 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

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3 years ago
MA of the first class lever may be equal to, greater than 1.Why​
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3 years ago
A bullet with kinetic energy of 400J strikes a wooden block where a 8.00x10N resistive force stops the bullet what is the penetr
tia_tia [17]

The "penetration of the bullet" is 5 m

<u>Explanation</u>:

A "bullet" with "kinetic energy" of = 400J

A resistive force stops the bullet  = 8.00 x 10 N

Work = change in energy  

Work = ∆ Kinetic Energy   (equation 1)

Work = F\times d   (equation 2)

From equations 1 and 2 we have,

F\times d = ∆ Kinetic Energy

Where ,

Kinetic Energy = 400 J

F = 8.00 x 10 N

(8.00 x 10 N) d = 400 J

(80 N) d = 400 J

d=\frac{400}{80}

d = 5 m

The penetration of the bullet is 5 m

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3 years ago
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